Math 25: Advanced Calculus
UC Davis, Spring 2011
Math 25 – Solutions to Homework Assignment #6
1. Let sequences (
s
n
)
∞
n
=1
and (
t
n
)
∞
n
=1
converge to respective limits
S
and
T
.
Fix some
>
0.
By definition, there exist numbers
m
s
, m
t
such
that for any
n
s
≥
m
s
and
n
t
≥
m
t
,

s
n
s

S

<
2
and

t
n
t

T

<
2
. If
n
≥
max
{
m
s
, m
t
}
, by the triangle inequality,

(
s
n

t
n
)

(
S

T
)

=

(
s
n

S
)

(
t
n

T
)
 ≤ 
s
n

S

+

t
n

T

<
2
+
2
=
. By definition,
then,
s
n

t
n
→
S

T
as
n
→ ∞
.
2. For (
s
n
)
∞
n
=1
and (
t
n
)
∞
n
=1
such that
s
n
=
√
n
+ 1
,
t
n
=
√
n,
Both sequences diverge to infinity, and hence the reasoning in problem
1. does not apply. However, for any
n
, since
a

b
= (
√
a

√
b
)(
√
a
+
√
b
),
s
n

t
n
=
√
n
+ 1

√
n
=
(
n
+ 1)

n
√
n
+ 1 +
√
n
=
1
√
n
+ 1 +
√
n
Now, since
0
≤
1
√
n
+ 1 +
√
n
≤
1
√
n
And because
1
√
n
→
0 as
n
→ ∞
, by the squeeze theorem, (
s
n

t
n
)
∞
n
=1
converges to 0.
3.
(a) lim
n
→∞
1
n
3
= 0, since
1
n
3
≤
1
n
, and
1
n
→
0 as
n
→ ∞
.
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 Winter '11
 Wiley
 Differential Equations, Calculus, Equations, Limits

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